So for your function, we are looking for the # x # values such that # G'(x) = 0 #. This is # (2x-1)/(3(x^2-x)^(2/3)) = 0 #. Now the only way this is possible is if # 2x-1 = 0 #. This gives # x = 1/2 #.

But we must also see where # G'(x) # is undefined!
Now we can't have a zero denominator, so # G'(x) # is undefined if # x^2-x = 0#. Factoring, we get # x(x-1) = 0 #. This gives # x = 0 # and # x=1 #.